Is the product of two Luzin-N functions still a Luzin-N function? Suppose $f$ and $g$ are two Luzin-N functions bounded on $[a,b]$.
Is the product of $f$ and $g$ still a Luzin-N function?
Known: if $f$, $g$ are absolutely continuous, then one may use the backward argument form Banach theorem and states that $f(x)g(x)$ had property.
(See: wiki's definition for Luzin-N https://en.wikipedia.org/wiki/Luzin_N_property)
However, I was thinking can we prove that the product of two Luzin-N functions is a Luzin-N function on $[a,b]$ pure directly and without the assumption that $f,g$ are absolutely continuous? and how many conditions can one drop?(absolutely continuous, continuous, bounded variation, bounded... etc.) Because, essentially, I was thinking that the proof of the statement is about the property of Lebesgue measure and Luzin-N, and should not be concerned by the continuity of $f$ and $g$.
I'm using the argument that $g(x)f(x)$ essentially is a metric scaling act on each other, thus if $|f|,|g|<M$ then $m(\{f(x)g(x):x\in E\})\leq M*(m(\{g(x):x\in E\})+m(\{f(x):x\in E\}))=0$. 
In the sense that: suppose $E$ a set covered by a sets of intervals $I_n$, the left product by a product $a\cdot E$ would scale the cover $I_n$ of $E$ by $a\cdot I_n$. Thus $m(a \cdot E)=a \cdot m(E)$. [However, in order to implement the above statement with $M$, this seem to be suggesting the usage of continuity of $f(x)g(x)$(which we can allow).] Thus no matter how $g\wedge f(E)$ would map the $E$, the scaling is less than $m(g(M\cdot E))=M\cdot m(g(E))$ for some large $M$ by fixing $g(E)$ and consider $f$ in the simple form $\varphi=\sum_i \alpha_i \chi_{E_i}$. Thus taking the limit in sup, one may obtain the argument.  Then one may reproduce the above argument and do it for two copies of the original sets, which is definitely larger.
Notice this argument used the fact that at least one of $f(x)$ or $g(x)$ is bounded $m$ almost everywhere. So, can one prove the statement without the usage of bounded at all, and where does continuity come to play in the proof?
I think that's enough, but I'm not sure if I still need to add the condition of $f,g$ being continuous into the argument. 
 A: This is an attempted proof, please help me to verify.
The product of two bounded Luzin-N function still a Luzin-N function.
Suppose $f$ and $g$ were two Luzin-N function on $[a,b]$ and $|f|,|g|<M$.
Basic 1:
For an interval $I$ and constant $a$, $m(a\cdot I)=a\cdot I$ i.e. the left product of $a$ on the elements in $I$ acted as a scaling factor for measure on $a\cdot I$. Further, the sign of $a$ doesn't matter, i.e. $m(a\cdot I)=|a|\cdot I$ if $a<0$.(assume $a\neq 0$, because $a=0$ is only going to make the proof easier as we are approximate by $\leq $.)
Suppose $E$ a set in $\Sigma$ and $m(E)<\infty$ exists. Then $m=\sup\{ \sum_n I_n: E\in \bigcup_n I\}$.
Here we notice that $m(a\cdot E)=|a|\cdot m(E)$ as usual.
Argument 2:
We extend the above observation in the following form. Notice that, if we fix $g(x)$, then the left product of  $f(x)$ on $g(x)$ is essentially a scaling metric of $f(x)$ acting on $g(x)$.
To formalize the argument, we notice that since $f(x)$ a measurable function bounded $m$ almost everywhere, it may be approximated (from below) by $\varphi_m=\sum_i^m \alpha_i\chi_{E_i}$, where $|\alpha_i|\leq M$. (Better, it can even be approximated by a step function.)  
We emphasize again that we were interested on the measure of the set $m(f\wedge g(E))=m(\{f(x)g(x):x\in E\})$. Notice that $f(x)$'s scaling on the $g(x)$ is equivalent to the scaling to the interval contain $g(x)$, and 
$m(\varphi_m\wedge g(E))
=m(\{(\sum_i^m \alpha_i \chi_{E_i} )(x)\cdot g(x):x\in E\})$
Notice since $g(E)$ was covered by a sets of intervals $\{I_n\}$, the simple function $\varphi_m$ act on $g$ at point $x$ essentially scales the interval $I_x$ that contains $g(x)$, and the the re scaled $I_x$ was extend to a factor of at most  $\sup\{|\alpha_x|:E_i\cap g^{-1}(I_x)\neq 0\}$, where $g^{-1}I_x$ indicate the preimage of $I_x$ and is not necessarily a function.
which seem to imply (one may seem to conclude but is false)
$m(\{(\sum_i^m \alpha_i \chi_{E_i} )(x)\cdot g(x):x\in E\})
\leq (\inf\{ \sum_j^n (I_j\cdot \alpha_x):
\alpha_x= \sup\{|\alpha_i|: E_i\cap g^{-1}(I_j)\neq \emptyset \},
g(E)\in \bigcup_j^n I_j\})
\leq (\inf\{ \sum_j^n (I_j\cdot M):
g(E)\in \bigcup_j^n I_j \}) 
=(\inf\{ M\cdot \sum_j^n I_j:
g(E)\in \bigcup_j^n I_j \})
=M\cdot m(g(E))$.
Correction 3:
The above statement is false, because we haven't use the fact that $f$ had property $N$, and consider $f(x)+x$ where $f(x)$ is the cantor's function, and set $g(x)=1$. The reason it didn't work was because the effect of $g$ on $f$ was not counted.
When looking at the sets of intervals (covers) $u_m$ for $f$ and $v_m$ for $g$, we note that that the scaling is happening to both $u_m$ and $v_m$, that $f(x)g(x)$ extended both $v_m$ and $u_n$.
We thus construct $\varphi^f_m\wedge g$ and $f\wedge \varphi^g_m$. Compare to the $f\wedge g$. We note that if $I_x$ covers $f\wedge g$ at $x$, it's measure is smaller than the extended cover of $\varphi^f_m\wedge g$ plus the cover for $f\wedge \varphi^g_m$ at $x$.
This complete the required proofs in the form that 
$m(f\wedge g(E))
=\lim_{m,n\rightarrow \infty}m(\varphi_{m}^f\wedge g(E))\cup (f\wedge \varphi_n^g(E)))
=\lim_{m,n\rightarrow \infty}m(\varphi_{m}^f\wedge g(E)) + m(f\wedge \varphi_n^g(E))
\leq M\cdot m(g(E))+M\cdot m(f(E))$.
If $E$ is a sets of measure $0$, since $f,g$ had property $N$, $m(f\wedge g(E))\leq M\cdot m(g(E))+M\cdot m(f(E))=0$.
Thus $f\wedge g$ is a Luzin N function.
